How do I show something is pathwise connected?

  • Thread starter amanda_ou812
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In summary: X={(x,y):a<=x<=b, c<=y<=d} can be proven by contradiction. Assuming X is not pathwise connected, then for any two points (x, y) and (u, v) in X, there exists no continuous function f:[0,1]→X such that f(0)=(x, y) and f(1)=(u, v). This implies that X is disconnected, which contradicts the fact that it is a subset of the connected set R^2. Therefore, X must be pathwise connected.
  • #1
I need to prove that X={(x,y):a<=x<=b, c<=y<=d}

I was thinking of using proof by contradiction.

Assume that X is not pathwise connected, then for a,b in X there is no continuous function that connects the two.

I can show that then the set is disconnected but not sure where to go after that.

Am I going about it the correct way?

Thanks!
 
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  • #2
Why can't you just give the path that connects two points??

Given (x,y) and (u,v). What path can you take that connects (x,y) and (u,v)?
 
  • #3
if I take a straight line path from the points, it would be the distance between the two, so sqrt((x-u)^2 + (y-v)^2)
 
  • #4
I don't care about the distance. I want a function [itex][0,1]\rightarrow X[/itex]. A linear path is ok. How do you write that in function language?
 
  • #5
i am not sure what you mean by function language but I would want f(0) = (x,y) and f(1) = (u, v)
 
  • #6
Xhat is the equation of the line connecting your two points??
 
  • #7
i want to say something like

given f(0) = (x, y) and p in [0,1] then f(p) = ((y-v)/(x-u)) * p + (x, y)

but that's not correct bc i am switching between R and R^2

do i need to make a continuous function for each component?
 
  • #8
Yes, you need to go linearly drom x to u and from y to v.
 
  • #9
do I do it peicewise? like f(p) = (x,y) for p=0, (u, v) for p=1 and then this other functions that I cannot figure out how to get.
 
  • #10
amanda_ou812 said:
I need to prove that X={(x,y):a<=x<=b, c<=y<=d}
You need to prove what about this set? That it is connected? Or that it is pathwise connected? A set can be "connected" without being "pathwise connected".

I was thinking of using proof by contradiction.

Assume that X is not pathwise connected, then for a,b in X there is no continuous function that connects the two.

I can show that then the set is disconnected but not sure where to go after that.

Am I going about it the correct way?

Thanks!
 
  • #11
Pathwise connected
 

1. What is pathwise connectedness?

Pathwise connectedness is a topological property that describes the ability to continuously deform one point in a space to another point in the same space while remaining within the space. In simpler terms, it means that any two points in the space can be connected by a continuous curve.

2. How do I prove that a space is pathwise connected?

To show that a space is pathwise connected, you need to demonstrate that for any two points in the space, there exists a continuous function that maps an interval between the two points onto a subset of the space. This function should also satisfy the condition that the starting and ending points of the interval are mapped to the two given points.

3. What is the difference between pathwise connectedness and connectedness?

Pathwise connectedness is a stronger property than connectedness. While connectedness only requires that a space cannot be separated into two disjoint open sets, pathwise connectedness additionally requires that any two points in the space can be connected by a continuous curve.

4. Can a disconnected space also be pathwise connected?

No, a disconnected space cannot be pathwise connected. If a space is disconnected, it means that it can be separated into two disjoint open sets. Therefore, any two points in the space cannot be connected by a continuous curve.

5. How is pathwise connectedness related to the fundamental group?

Pathwise connectedness is a necessary condition for the fundamental group of a space to be nontrivial. In other words, if a space is not pathwise connected, its fundamental group will always be trivial (containing only the identity element). However, the converse is not true, as there are pathwise connected spaces with trivial fundamental groups.

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